• C

getbuf() in C


char *a, b, *c;
a=getbuff();
b=*a;
c=otherfunction(a);
...
char *getbuff()
{
     char buff[8];
     ...
     return (char*)buff;
}

After executed, what will be in variable "b"? And under what conditions might this not be so?
Are there any way to wirte an alternative, but equivalent, as "char *getbuff()"?
DecaiAsked:
Who is Participating?
 
Sjef BosmanConnect With a Mentor Groupware ConsultantCommented:
I think we'd better make it good practice to never supply the full answer, or throw in some pitfalls for the cut/paste student. I've seen lots of complete code passing by, to be called "homework" as I've noticed, but this was an inconspicuous one, so "I plead guilty". I think it's not illegal EE behaviour though, or is it?

I suppose we're all too happy to be of help. I hope the others will be more hesitant to give very detailed answers to questions like these.

Thanks for the warning!
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skypalaeCommented:
Hi Decai,

since buff[8] is statically allocated inside the getbuff() the pointer becomes non-valid after return. Accessing the memory under that pointer (the b=*a line) will most probably cause the program crash (also passing a non-valid pointer to otherfunction() will most probably cause the crash)

if you want to make the pointer valid as long as you wish use malloc() and free() functions instead.

char *getbuff () {
    char *buff = (char *) malloc (sizeof(char) * 8) ;
    ...
    return buff ;
}

but don't forget to free() the memory at the end of program or you'll get memory leaks.

Cheers! S.
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skypalaeCommented:
Decai,

and to answer you completely .. the 'b' will contain a[0] element.

S.
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Sjef BosmanGroupware ConsultantCommented:
In my humble opinion
1) the content of variable b can be predicted, but it is higly unsafe!
2) this might not be predictable when you have one or more functions that catch a signal() in Linux;
  if you use *a in a later stage, after having called one or more other functions, *a might have a totally different value
3) for an alternative, see above (malloc()) but don't forget to free() the memory once unnecessary (there is no garbage collection in C)

Unlike skypalae says, the buff[8] isn't statically allocated, it is on the stack. There are 3 types of memory: main, stack and heap. Static variables are in main memory. Your buff is on the stack, and when the function getbuff returns, stack content gets lost. The stack's memory probably still exists, so you can still "use" it. It's memory isn't it, and it's yours to use. Fiddling with the stack is very dangerous though, your program might crash! The function malloc() dynamically allocates space (from the heap). It is very good practice to free() the memory allocated by malloc() once done with it, to keep your program from growing out of bounds.
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CodeDeamonCommented:
This is a test question if I have ever seen it.  I love it when "programmers" try to get actual programmers to do their work for them.
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